The masses of two fixed spheres are M and 2M | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let the masses be $M_1 = M$ and $M_2 = 2M$. The distance between their centres is $d = 10R$. The particle of mass $m = \frac{M}{10}$ is projected

Vedclass · Accepted Answer

$\sqrt{\frac{6 GM}{5 R}}$

Vedclass · Answer

(D) Let the masses be $M_1 = M$ and $M_2 = 2M$. The distance between their centres is $d = 10R$. The particle of mass $m = \frac{M}{10}$ is projected from the mid-point,which is at a distance $r_1 = 5R$ from $M_1$ and $r_2 = 5R$ from $M_2$.By the law of conservation of energy,the total energy at the mid-point must equal the total energy at infinity (where potential energy and kinetic energy are zero).Initial Energy $E_i = K_i + U_i = \frac{1}{2}mv^2 - \frac{GM_1m}{r_1} - \frac{GM_2m}{r_2}$.Substituting the values: $E_i = \frac{1}{2}mv^2 - \frac{GMm}{5R} - \frac{G(2M)m}{5R} = \frac{1}{2}mv^2 - \frac{3GMm}{5R}$.Final Energy $E_f = 0$.Setting $E_i = E_f$: $\frac{1}{2}mv^2 = \frac{3GMm}{5R}$.Solving for $v$: $v^2 = \frac{6GM}{5R}$,so $v = \sqrt{\frac{6GM}{5R}}$.

Similar Questions

$A$ body is projected vertically from the surface of the earth of radius $R$ with a velocity equal to half of the escape velocity. The maximum height reached by the body is

The escape velocity on Earth is $11.2 \, km/s$. On another planet having twice the radius and $8$ times the mass of the Earth,the escape velocity will be ......... $km/s$.

The magnitude of potential energy per unit mass of an object at the surface of the Earth is $E$. Then,the escape velocity of the object is ..........

$A$ ball $A$ of mass $m$ falls to the surface of the earth from infinity. Another ball $B$ of mass $2m$ falls to the earth from a height equal to six times the radius of the earth. The ratio of the velocities of $A$ and $B$ on reaching the earth is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module